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Mathematics Question on Differential Equations

The particular solution of the differential equation (y−x2)dy=(1−x3)dx(y - x^2) dy = (1 - x^3) dx with y(0)=1y(0) = 1, is:

A

y2=x2+2log⁡e∣1+x∣+1y^2 = x^2 + 2 \log_e |1 + x| + 1

B

y2=1+x2+2log⁡e∣1+x2∣y^2 = 1 + x^2 + 2 \log_e \left| \frac{1 + x}{2} \right|

C

y2=x2+2x−3y^2 = x^2 + 2x - 3

D

y2=x2+2x+1y^2 = x^2 + 2x + 1

Answer

y2=x2+2log⁡e∣1+x∣+1y^2 = x^2 + 2 \log_e |1 + x| + 1

Explanation

Solution

Rewrite the given differential equation as:

dydx=1−x3y−x2y.\frac{dy}{dx} = \frac{1 - x^3}{y - x^2 y}.

Factor yy in the denominator:

dydx=1−x3y(1−x2).\frac{dy}{dx} = \frac{1 - x^3}{y(1 - x^2)}.

Separate variables:

y(1−x2)dy=(1−x3)dx.y(1 - x^2) dy = (1 - x^3) dx.

Integrate both sides:

∫y dy=∫1−x31−x2dx.\int y \, dy = \int \frac{1 - x^3}{1 - x^2} dx.

The left-hand side integrates to:

∫y dy=y22.\int y \, dy = \frac{y^2}{2}.

Simplify the right-hand side using partial fractions:

1−x31−x2=1+x.\frac{1 - x^3}{1 - x^2} = 1 + x.

Thus:

∫(1+x)dx=x+x22.\int (1 + x) dx = x + \frac{x^2}{2}.

Equating both sides:

y22=x+x22+C.\frac{y^2}{2} = x + \frac{x^2}{2} + C.

Multiply through by 2:

y2=x2+2x+2C.y^2 = x^2 + 2x + 2C.

Using the initial condition y(0)=1y(0) = 1:

12=0+0+2C  ⟹  C=12.1^2 = 0 + 0 + 2C \implies C = \frac{1}{2}.

Thus, the solution is:

y2=x2+2x+1.y^2 = x^2 + 2x + 1.